2.4 Equations and Inequalities with Absolute Values

2.4 Equations and Inequalities with Absolute Values

Absolute value expressions are frequently tested on the Bocconi Admission Test, appearing both as standalone algebra problems and within function domain or calculus questions. Because modulus operations are piecewise non-linear, standard algebraic manipulation often introduces extraneous solutions or sign errors unless candidate domains are partitioned rigorously.

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Definition and Geometric Interpretation

Formal Algebraic Definition

The absolute value of a real number $x$ is defined piecewise: $|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$ The output is always non-negative: $|x| \ge 0$ for all $x \in \mathbb{R}$.

Geometric Distance on the Real Line

Geometrically, $|x|$ is the Euclidean distance from coordinate $x$ to the origin $0$. For any two coordinates $x$ and $a$, the distance between them is: $\text{Distance}(x, a) = |x - a|$ This geometric viewpoint solves many problems by inspection:

• $|x - 3| = 5$: Points $x$ at distance $5$ from $3$, yielding $3 + 5 = 8$ and $3 - 5 = -2$.
• $|x - 4| < 2$: Points $x$ within distance $2$ of $4$, yielding open interval $(4 - 2, 4 + 2) = (2, 6)$.

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Fundamental Properties of Modulus

For all real numbers $x, y$:

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Absolute Value Equations: Solution Strategies

Form 1: $|f(x)| = c$ (Constant Right Side)

• If $c < 0$: No solution ($S = \emptyset$).
• If $c = 0$: Single linear equation $f(x) = 0$.
• If $c > 0$: Disjunction of two branches: $f(x) = c$ or $f(x) = -c$.

Form 2: $|f(x)| = g(x)$ (Variable Right Side)

Because the left side is non-negative, real solutions require the existence constraint $g(x) \ge 0$. Under this condition, split into $f(x) = g(x)$ or $f(x) = -g(x)$. Any algebraic candidate causing $g(x) < 0$ is extraneous and must be rejected.

Form 3: $|f(x)| = |g(x)|$ (Modulus on Both Sides)

Both sides are non-negative, so no extraneous sign contradictions arise. The equation is directly equivalent to $f(x) = g(x)$ or $f(x) = -g(x)$, or by squaring: $[f(x) - g(x)][f(x) + g(x)] = 0$.

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Absolute Value Inequalities: Conjunction vs. Disjunction

For any positive constant $c > 0$:

Bounded Form $|f(x)| < c$ or $\le c$ (Conjunction / AND)

Confines $f(x)$ within a single bounded interval: $-c < f(x) < c \quad (\text{or } -c \le f(x) \le c)$

Unbounded Form $|f(x)| > c$ or $\ge c$ (Disjunction / OR)

Pushes $f(x)$ into two disjoint unbounded rays: $f(x) < -c \quad \text{or} \quad f(x) > c$ The solution is the union of two intervals: $(-\infty, x_1) \cup (x_2, \infty)$.

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Multi-Absolute-Value Expressions and Interval Decomposition

For equations with multiple modulus terms, apply the Method of Sub-Intervals:

1. Set each modulus argument to zero to identify critical points: $r_1 < r_2 < \dots < r_n$.
2. Partition the real line into $n + 1$ sub-intervals.
3. Replace each modulus term with $+(\dots)$ or $-(\dots)$ based on its sign in that interval.
4. Solve the algebraic equation on each sub-interval.
5. Intersect candidate solutions with the active sub-interval; discard any lying outside.
6. Take the set union of all valid solutions.

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Worked Step-by-Step Examples

Example 1: Variable Right-Hand Side with Extraneous Root

Solve $|2x - 5| = 3x - 10$ for real $x$:

1. State existence condition: $3x - 10 \ge 0 \implies x \ge 10/3 \approx 3.33$.
2. Branch 1: $2x - 5 = 3x - 10 \implies -x = -5 \implies x = 5$. Since $5 \ge 10/3$, $x = 5$ is valid.
3. Branch 2: $2x - 5 = -(3x - 10) \implies 5x = 15 \implies x = 3$. Since $3 = 9/3 < 10/3$, $x = 3$ is extraneous.
4. Check: At $x = 5$, $|2(5) - 5| = 5 = 3(5) - 10$. Valid! Solution set: $S = {5}$.

Example 2: Bounded Modulus Inequality and Integer Counting

Solve $|4 - x| \le 6$ and count integer solutions:

1. By symmetry: $|x - 4| \le 6$.
2. Double inequality: $-6 \le x - 4 \le 6$.
3. Add 4: $-2 \le x \le 10$, giving closed interval $[-2, 10]$.
4. Integer count: $10 - (-2) + 1 = 13$ integers.

Example 3: Multi-Modulus Equation

Solve $|x + 2| + |x - 4| = 8$:

1. Critical points: $x = -2$ and $x = 4$, creating three intervals: $(-\infty, -2)$, $[-2, 4]$, and $(4, \infty)$.
2. Interval $x < -2$: $-(x + 2) - (x - 4) = 8 \implies -2x + 2 = 8 \implies x = -3$ (in interval, valid).
3. Interval $-2 \le x \le 4$: $(x + 2) - (x - 4) = 8 \implies 6 = 8$ (false, no solution).
4. Interval $x > 4$: $(x + 2) + (x - 4) = 8 \implies 2x - 2 = 8 \implies x = 5$ (in interval, valid).
5. Solution set: $S = {-3, 5}$. Solution sum: $-3 + 5 = 2$.